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              <h1>什么是锁膨胀升级 - 交互式测验</h1>
              <p>检验你对文档内容的理解程度</p>
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      <script>
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  {
    "question": "根据文档，Java中的“锁膨胀升级”过程的正确顺序是什么？",
    "options": [
      "轻量级锁 -> 偏向锁 -> 重量级锁",
      "偏向锁 -> 轻量级锁 -> 重量级锁",
      "自旋锁 -> 互斥锁 -> 重量级锁",
      "无锁 -> 偏向锁 -> 重量级锁"
    ],
    "correctAnswerIndex": 1,
    "explanation": "锁膨胀升级是一个单向过程，旨在根据竞争的激烈程度来平衡性能和开销。它始于最低开销的偏向锁，当出现竞争时升级为轻量级锁，在竞争进一步加剧时，最终升级为开销最大但能处理高并发的重量级锁。",
    "source": "文档中「锁膨胀升级的定义」-「1. 锁的状态和升级路径」部分",
    "hint": "文档中提到，JVM为锁设计了三种状态以适应不同的并发环境，这个过程始于一种为优化单线程访问而设计的状态。"
  },
  {
    "question": "当一个线程尝试获取一个已经被其他线程持有的轻量级锁，并且经过多次自旋后仍然失败，接下来最可能发生什么情况？",
    "options": [
      "锁降级为偏向锁，以减少开销。",
      "线程将无限期地自旋，直到获取锁为止。",
      "锁膨胀为重量级锁，当前线程将被阻塞。",
      "JVM将抛出并发修改异常 (ConcurrentModificationException)。"
    ],
    "correctAnswerIndex": 2,
    "explanation": "轻量级锁通过自旋来避免线程阻塞，但如果自旋在一定次数内无法获得锁，说明竞争已经很激烈。此时JVM会认为继续自旋是在浪费CPU资源，因此会将锁升级为重量级锁，并利用操作系统的同步机制来阻塞当前线程，从而避免CPU空转。",
    "source": "文档中「锁膨胀升级的原因」-「5. 锁撤销和自旋策略」部分",
    "hint": "当自旋策略在高竞争下失效时，JVM会采取一种更传统但开销更大的方式来管理线程，以避免长时间占用CPU资源。"
  },
  {
    "question": "下列哪项是导致偏向锁（Biased Locking）升级为轻量级锁（Lightweight Locking）的直接触发条件？",
    "options": [
      "同一个线程递归地获取该锁。",
      "系统内存资源不足。",
      "持有偏向锁的线程执行时间过长。",
      "另一个线程尝试获取该偏向锁。"
    ],
    "correctAnswerIndex": 3,
    "explanation": "偏向锁的核心思想是，锁在大多数情况下由同一个线程获取。这个“偏向”的假设一旦被打破——即有第二个线程来竞争这个锁——偏向锁就会被撤销，并升级为轻量级锁以处理这种新的竞争情况。",
    "source": "文档中「锁膨胀升级的定义」-「1. 锁的状态和升级路径」部分",
    "hint": "偏向锁的设计初衷是优化“无竞争或极低竞争的场景”。当什么情况发生时，这个前提就不成立了？"
  },
  {
    "question": "文档中提到的哪种锁类型，在等待锁时不会立即阻塞线程，而是让线程持续占用CPU进行“忙等待”？",
    "options": [
      "重量级锁 (Heavyweight Lock)",
      "读写锁 (ReadWriteLock)",
      "自旋锁 (Spinlock)",
      "公平锁 (Fair Lock)"
    ],
    "correctAnswerIndex": 2,
    "explanation": "自旋锁的核心机制是在获取锁失败时，线程不会被挂起（阻塞），而是执行一个空的循环（自旋），不断尝试获取锁。这种方式适用于锁被持有时间极短的场景，可以避免线程上下文切换带来的开销。",
    "source": "文档中「锁的基本概念」-「2. 锁的基本类型」部分",
    "hint": "这种锁的策略适用于锁等待时间很短的场景，以避免线程的频繁上下文切换。"
  },
  {
    "question": "根据文档，锁膨胀为重量级锁后，会对系统性能产生什么主要负面影响？",
    "options": [
      "增加CAS操作失败的概率，浪费CPU资源。",
      "导致内存泄漏，因为锁对象无法被回收。",
      "依赖操作系统进行线程调度，导致昂贵的上下文切换。",
      "简化了并发编程模型，但降低了代码可读性。"
    ],
    "correctAnswerIndex": 2,
    "explanation": "当锁升级为重量级锁时，其实现依赖于操作系统的互斥量（Mutex）。线程在获取锁失败时会被操作系统挂起（阻塞），并在锁被释放时被唤醒。这个阻塞和唤醒的过程涉及从用户态到内核态的转换以及线程上下文切换，这是非常耗费性能的操作。",
    "source": "文档中「锁膨胀升级的影响」-「1. 性能下降」-「上下文切换的增加」部分",
    "hint": "当锁的竞争从用户态转移到内核态时，会引入一种代价高昂的操作。"
  },
  {
    "question": "在“读多写少”的高并发场景中，为了减少锁竞争并避免锁膨胀，文档推荐使用哪种优化策略？",
    "options": [
      "使用粗粒度锁，将所有读写操作都包含进去。",
      "禁用偏向锁，直接从轻量级锁开始。",
      "使用读写锁（ReentrantReadWriteLock）。",
      "增加线程池的大小以提高并发度。"
    ],
    "correctAnswerIndex": 2,
    "explanation": "读写锁允许多个线程同时进行读操作，只有在进行写操作时才会互斥。这非常适合“读多写少”的场景，因为它极大地提高了读操作的并发性，减少了因读操作而产生的锁竞争，从而有效避免了不必要的锁膨胀。",
    "source": "文档中「锁膨胀升级的优化策略」-「2. 优化锁的使用方式」部分",
    "hint": "文档中提到一种特殊的锁机制，它允许多个线程同时读共享资源，但在写操作时只允许一个线程操作。"
  },
  {
    "question": "以下哪项不是文档中提到的导致锁膨胀升级的直接原因？",
    "options": [
      "线程竞争激烈，大量CAS操作失败。",
      "锁的粒度设计不合理，例如使用了粗粒度锁。",
      "垃圾回收（GC）活动过于频繁。",
      "线程在持有锁的情况下进行递归调用。"
    ],
    "correctAnswerIndex": 2,
    "explanation": "文档中详细列举了导致锁膨胀的原因，包括线程竞争、线程数增加、锁粒度不合理、递归调用等。虽然频繁的GC可能会影响应用性能，甚至触发STW（Stop-The-World），但它并不是锁状态升级的直接触发机制。",
    "source": "文档中「锁膨胀升级的原因」部分",
    "hint": "请回顾文档中专门论述锁膨胀原因的章节，并找出那个未被提及的因素。"
  },
  {
    "question": "为了从根本上避免锁竞争和膨胀问题，文档提倡的一种高级并发策略是什么？",
    "options": [
      "总是使用重量级锁以保证绝对的线程安全。",
      "采用无锁数据结构和无锁编程范式。",
      "将所有共享资源设置为 final。",
      "为每个线程创建独立的锁对象。"
    ],
    "correctAnswerIndex": 1,
    "explanation": "无锁编程是一种通过CAS等原子操作来保证线程安全的技术，它完全避免了传统锁机制的使用。因此，它从根本上消除了锁竞争、死锁以及锁膨胀等问题。例如，`java.util.concurrent`包中的`ConcurrentLinkedQueue`就是一种无锁数据结构。",
    "source": "文档中「锁膨胀升级的优化策略」-「3. 采用无锁数据结构」部分",
    "hint": "这种策略通过避免使用锁，而是依赖于CAS等原子操作来实现线程安全。"
  }
];
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        }

        function showFinalScore() {
            document.querySelector('.quiz-body').style.display = 'none';
            document.getElementById('quiz-footer').style.display = 'none';

            const correctCount = score;
            const totalQuestions = quizData.length;
            const answeredCount = userAnswers.filter(a => a !== null).length;
            const wrongCount = answeredCount - correctCount;
            const skippedCount = totalQuestions - answeredCount;
            const accuracy = totalQuestions > 0 ? Math.round((correctCount / totalQuestions) * 100) : 0;
            const errorRate = totalQuestions > 0 ? Math.round((wrongCount / totalQuestions) * 100) : 0;

            const finalScoreHtml =
                '<div class="final-score-container">' +
                '<h2>答题报告</h2>' +
                '<div class="score-summary">' +
                '<div class="score-card">' +
                '<h3>得分</h3>' +
                '<p>' + correctCount + ' / ' + totalQuestions + '</p>' +
                '</div>' +
                '<div class="score-card">' +
                '<h3>正确率</h3>' +
                '<p>' + accuracy + '%</p>' +
                '</div>' +
                '<div class="score-card">' +
                '<h3>错误率</h3>' +
                '<p>' + errorRate + '%</p>' +
                '</div>' +
                '<div class="score-card score-details">' +
                '<table>' +
                '<tr><td>正确</td><td>' + correctCount + '</td></tr>' +
                '<tr><td>错误</td><td>' + wrongCount + '</td></tr>' +
                '<tr><td>未答</td><td>' + skippedCount + '</td></tr>' +
                '</table>' +
                '</div>' +
                '</div>' +
                '<div class="report-actions">' +
                '<button id="review-btn">回顾测验</button>' +
                '<button class="primary" onclick="location.reload()">再试一次</button>' +
                '</div>' +
                '</div>';

            let reportEl = document.querySelector('.final-score-container');
            if (reportEl) {
                reportEl.remove();
            }
            quizContentEl.insertAdjacentHTML('beforeend', finalScoreHtml);
            document.getElementById('review-btn').addEventListener('click', reviewQuiz);
        }

        function reviewQuiz() {
            const reportEl = document.querySelector('.final-score-container');
            if (reportEl) {
                reportEl.remove();
            }

            document.querySelector('.quiz-body').style.display = 'block';
            document.getElementById('quiz-footer').style.display = 'flex';
            
            currentQuestionIndex = 0;
            loadQuestion();
            updateNavigationButtons(); // Ensure nav buttons are correctly displayed
        }

        // Event Delegation for options
        optionsListEl.addEventListener('click', (event) => {
            if (event.target && event.target.matches('li.option-item')) {
                // Check if the question has already been answered
                if (userAnswers[currentQuestionIndex] === null) {
                    const selectedIndex = parseInt(event.target.dataset.index, 10);
                    selectOption(selectedIndex);
                }
            }
        });

        prevBtn.addEventListener('click', goToPreviousQuestion);
        nextBtn.addEventListener('click', goToNextQuestion);
        hintBtn.addEventListener('click', showHint);
        
        // 初始加载
        loadQuestion();
      </script>
    </body>
    </html>
  